Hardest Integral you will ever do!!!


by nshiell
Tags: hardest, integral
nshiell
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#1
May12-08, 07:51 PM
P: 9
int(exp(-a*r^2+a*r*x*cos(theta))*r))

This is a double integral over a keystone shape in polar coordinates. (so limits are finite, and not a complete circle).

I've tried all sorts of things. I am not even sure if there is an analytic solution. Is there away to show there isn't one?
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Defennder
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#2
May12-08, 09:54 PM
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What is the order of integration? With respect to r first, then theta, or the other way round? And what is x here? Isn't this supposed to be in polar coordinates?
Dick
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#3
May12-08, 11:01 PM
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Quote Quote by nshiell View Post
int(exp(-a*r^2+a*r*x*cos(theta))*r))

This is a double integral over a keystone shape in polar coordinates. (so limits are finite, and not a complete circle).

I've tried all sorts of things. I am not even sure if there is an analytic solution. Is there away to show there isn't one?
If you tried all sorts of things, can you show us one of them?

nshiell
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#4
May13-08, 08:30 AM
P: 9

Hardest Integral you will ever do!!!


Okay so first the x is just another constant, I suppose it was a poor choice. And the order doesn't matter, which ever makes it easiest. And second, as for attempts i have made. The terms in the exponential are the result of a vector addition because the formal has to do with the distance from the point at distance x from the origin making an angle theta with the point on the keystone shape. This perspective makes the limits of integration easier since the origin is the origin of the keystone. What I tried was to integrate using the point at x as the origin. Then use cartestian coordinates. Integrating first with respect to y with limits as functions of x (to circles for the upper and lower arcs of the keystone and 2 lines as the other edges) however this gave me just as ugly an express.
Dick
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#5
May13-08, 09:21 AM
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Quote Quote by nshiell View Post
Okay so first the x is just another constant, I suppose it was a poor choice. And the order doesn't matter, which ever makes it easiest. And second, as for attempts i have made. The terms in the exponential are the result of a vector addition because the formal has to do with the distance from the point at distance x from the origin making an angle theta with the point on the keystone shape. This perspective makes the limits of integration easier since the origin is the origin of the keystone. What I tried was to integrate using the point at x as the origin. Then use cartestian coordinates. Integrating first with respect to y with limits as functions of x (to circles for the upper and lower arcs of the keystone and 2 lines as the other edges) however this gave me just as ugly an express.
Maybe it would be better if you could post a description of the original problem? The integral does look pretty funny.
Defennder
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#6
May13-08, 10:03 AM
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I just tried it at integrals.wolfram.com Turns out that either case you don't get an elementary function. Integrating wrt r first gives something with erf, the other one had no solution.
nshiell
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#7
May13-08, 10:51 AM
P: 9
Okay here is the original problem. A cluster of charge is deposited onto a conductive surface and disperse like,

p(r,t) = N*q/[(2*pi)*(2ht+w^2)]*exp[(-r^2)/(2*(2ht+w^2)] (This is the charge density)

where h is 1/rc of the surface, w is the width of the cluster (unimportant here), N # of charges in cluster (unimportant), and q is electron charge. A distance d from the location of the charge a keystone shaped pad collects charge. I need to find the charge on the pad as a function of time. This is done my integrating the charge density equation over the pad.

Since w,h and t and not important rite now the equation looks better if a = 2ht+w^2

p(r,t) = N*q/[(2*pi)*a]*exp[(-r^2)/(2*a)]
Defennder
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#8
May13-08, 11:06 AM
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r is a variable, so how is it justified you can substitute a = 2ht + w^w if h=1/rc? Where did theta come from?
Dick
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#9
May13-08, 11:19 AM
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Yes, so what is rc if h=1/rc? And to solve a problem like that, if the pad is 'collecting' charge you wouldn't just integrate the charge density over the pad. You would do that if the charge were just floating over the pad. Don't you want to integrate the current over the region of the pad boundary where it's flowing in?
nshiell
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#10
May13-08, 11:30 AM
P: 9
1/rc is the RC time constant of the surface material. And yeah your rite the charge flows over the pads, and the pads detect it capacitively.
Dick
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#11
May13-08, 01:31 PM
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Quote Quote by nshiell View Post
1/rc is the RC time constant of the surface material. And yeah your rite the charge flows over the pads, and the pads detect it capacitively.
Ok. That's looking reasonable. It's a normalized charge distribution. It's looking a lot less strange than the original integral. Now what is a 'keystone'? Is it just a section of an annulus? If that's the case then the integral is easy.
nshiell
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#12
May13-08, 01:57 PM
P: 9
yeah the keystone shape is a section of an annulus. Though the charge is not deposited at the origin of the annulus. its deposited at an arbitrary point.
Dick
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#13
May13-08, 02:43 PM
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Quote Quote by nshiell View Post
yeah the keystone shape is a section of an annulus. Though the charge is not deposited at the origin of the annulus. its deposited at an arbitrary point.
You shouldn't have any problem integrating p(r,t) over a 'keystone'. Just substitute u=r^2/(2a).
nshiell
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#14
May14-08, 12:39 PM
P: 9
yeah the r integral is not bad, but the theta integral is. Thats the part I can not get past.
Dick
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#15
May14-08, 12:54 PM
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Quote Quote by nshiell View Post
yeah the r integral is not bad, but the theta integral is. Thats the part I can not get past.
There's no theta in the integrand. Why is it hard?


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